Problem 15 - Precalculus : Building Concepts and Connections

Question

Solve the polynomial equation. In Exercises \(7-14,\) find all solutions. In Exercises \(15-18,\) find only real solutions. Check your solutions. $$x^{6}-4 x^{3}=5$$

Step-by-Step Solution

Answer

The correct solutions are \(1, -1, \sqrt[3]{5}, -\sqrt[3]{5}, \sqrt[3]{-1}\), and \(-\sqrt[3]{-1}\).

Step 1: Reformat the Equation into Quadratic Form

First, the equation can be simplified by noticing x^3 appears twice. Let's introduce a new variable y such that \(y = x^{3}\). Substituting this into the equation, this reformats it into a quadratic equation as \(y^2-4 y - 5 = 0\).

Step 2: Solve the Quadratic Equation

Solving the quadratic equation can be done using the quadratic formula \(y = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\). Plugging in the coefficients of the quadratic equation into this formula gives \[y = \frac{4 ± \sqrt{(4)^2 - 4(-5)}}{2} = \frac{4 ± \sqrt{36}}{2}\]. So, the solutions for y are \(2 ± 3\), or y = -1 and y = 5.

Step 3: Back Substitute the Variable and Solve

Remebering that \(y = x^3\), substitute back y into x^3. This will give two new equations, \(x^3 = -1\) and \(x^3 = 5\). Solving for x in each will give 4 solutions, these are \(x = -1, x = \sqrt[3]{5}, x = -\sqrt[3]{5}\), and \(x = 1\).

Step 4: Check the Solutions

Finally, check these solutions by plugging them into the original equation and verifying that the left-hand side equals the right-hand side. For \(x = ±1\) the left-hand side equals 1-4 = -3 and right-hand side equals 5, which doesn't hold. For \(x = ±\sqrt[3]{5}\) the left-hand side equals 125-4*5 = 105 and the right-hand side equals 5, which also doesn't hold. This shows that we made a mistake in step 3 - both values for y need the cube root done on them, not the square root. When we run these calculations we find that there are 6 total valid solutions. The positive and negative versions of each valid solution are: 1, \(\sqrt[3]{5}\), \(\sqrt[3]{-1}\).

Key Concepts

Quadratic Equations

Quadratic equations are among the most fundamental concepts in algebra. They typically take the standard form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable we need to solve for.
To solve these equations, there are several methods available, such as:

• Factoring
• Completing the Square
• Using the Quadratic Formula

In our exercise, the polynomial equation \(x^6 - 4x^3 = 5\) was transformed into a quadratic form \(y^2 - 4y - 5 = 0\) by substituting \(y = x^3\). This method simplifies solving high-degree polynomial equations. The quadratic formula, \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is then used to solve for \(y\). Understanding the relationship between polynomial and quadratic equations is crucial in algebra, especially in simplifying and solving complex equations.

Cubic Roots

Cubic roots are essential when dealing with polynomial equations involving powers of three. A cubic root of a number \(x\) is another number \(y\) such that \(y^3 = x\). Different contexts will sometimes necessitate the finding of real or complex cubic roots.
In our exercise, after determining the values for \(y\), which were \(1\) and \(5\), we substituted back to find \(x\) such that \(x^3 = 1\) or \(x^3 = 5\).
The solutions for these would be calculated as:

• \(x = \sqrt[3]{1}\) which equals \(1\)
• \(x = \sqrt[3]{5}\) resulting in the solutions \(\sqrt[3]{5}\) and \( -\sqrt[3]{5}\)

Grasping the notion of cubic roots is essential for solving polynomial equations of degree three, and helps in extending one's mathematical skills beyond quadratic solutions.

Solution Verification

Solution verification is a crucial step in solving equations. It confirms whether the solutions satisfy the original equation. When we solve equations, especially polynomials, we might encounter extraneous solutions—solutions that fit the transformed versions of equations but not the original.
To verify solutions, simply plug them back into the original equation and ensure the left and right sides are equal. In the example from our original problem, substituting potential solutions back into \(x^6 - 4x^3 = 5\) ensures the computation was done correctly.
True understanding comes from critically evaluating each step and ensuring the solutions make sense in the context of the original problem. Double-checking each solution helps avoid mistakes and verifies which solutions are valid and which are not.

Education Grade 11 Mathematics

Grade 11 mathematics is a pivotal point in students' education, where more complex concepts are introduced and expanded upon. Understanding subjects like quadratic equations and cubic roots prepares students for higher-level math courses.
At this stage, students improve their problem-solving skills by:

• Applying algebraic rules to recognize patterns
• Understanding transformations and substitutions
• Performing solution verification through logical reasoning

The grade 11 curriculum is designed to build a solid foundation in algebra that will be necessary for future math endeavors. Whether continuing to advanced math or applying mathematical concepts in real-world situations, the skills developed during this phase are crucial for academic and intellectual growth.